如何捕捉json.decoder.JSONDecodeError？

Question

我的程序没有捕获json json.decoder.JSONDecodeError，尽管我已经编写了捕获这些错误的代码(参见try-except block)。我做错了什么？

错误码：

raise JSONDecodeError("Expecting value", s, err.value) from None
json.decoder.JSONDecodeError: Expecting value: line 1 column 1 (char 0)

我的代码：

def check_code():
    url = get_url()
    headers = {'Accept': 'application/json'}
    response = requests.get(url, headers=headers).json()
    print(response)
    try:
        if (account := response.get('account')) and account.get('active'):
            status = "active account "

        else:
            status = "not active account "
    except (JSONDecodeError, json.JSONDecodeError):
        pass

    return status

我随机得到DecodeError，有时在2-3次尝试之后，有时在12-13次之后

Answer 1

因为您需要在将响应转换为json时放置try子句，而不是在读取时。

使用下面的代码：

def check_code():
    url = get_url()
    headers = {'Accept': 'application/json'}
    try:
        response = requests.get(url, headers=headers).json()
        print(response)
    except (JSONDecodeError, json.JSONDecodeError):
        pass

    if (account := response.get('account')) and account.get('active'):
        status = "active account "
    else:
        status = "not active account "

    return status